Figure 8 shows the distance-time graph for a car travelling at 15 m/s - AQA - GCSE Physics Combined Science - Question 6 - 2018 - Paper 2

When the brakes are applied, the kinetic energy of the car is converted into heat energy through friction between the brake pads and the wheels. This process results in a decrease in the kinetic energy of the car, causing the vehicle to slow down. The energy conversion to heat raises the temperature of the brake components, thereby increasing their thermal energy.

To find the initial velocity of the lorry, we can use the equation of motion:

$v^2 = u^2 + 2as$

Where:

• $v$ is the final velocity (19 m/s)
• $u$ is the initial velocity
• $a$ is the acceleration (2.0 m/s²)
• $s$ is the distance (84 m)

Rearranging and substituting the values:

$19^2 = u^2 + 2 imes 2 imes 84$

Calculating:

$361 = u^2 + 336$

Thus:

$u^2 = 361 - 336 = 25$

Taking the square root:

$u = 5 ext{ m/s}$

The initial velocity of the lorry is 5 m/s.

In Figure 9, we can observe the relationships between thinking distance, braking distance, and stopping distance as the speed of the car increases.

1. Thinking distance: This is directly proportional to the speed of the car; as speed increases, the distance the car travels during the driver's reaction time also increases.

2. Braking distance: This increases at an increasing rate with speed. This is due to the fact that at higher speeds, more kinetic energy is present, requiring greater stopping distance to come to a halt.

3. Stopping distance: This is the total distance a car travels from the moment the driver identifies the need to stop until the vehicle halts. It is the sum of thinking distance and braking distance, which means it also increases with speed.

Overall, all three distances increase with the speed of the vehicle, and the rate of this increase is greater for the braking distance as speed rises.